Answers Investigation 4

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1 Answers Applications 1 4. Patterns 2 and 4 can fold to form closed boxes. Patterns 1 and 3 cannot fold to form closed boxes. 10. Sketch of box and possible net: 5. a. Figures 1 and 2 can be folded to form a closed box. Pattern C cannot. b. Figure 1: 1 unit * 1 unit * 4 units Figure 2: 1 unit * 2 units * 4 units c. Figure 1: 18 sq. units Figure 2: 28 sq. units d. Figure 1: 4 cubes Figure 2: 8 cubes 6. a. 2 cm * 4 cm * 1 cm b. Possible answers: There are two of each of these faces: 2 cm * 3 cm (area is 6 cm 2 ); 2 cm * 5 cm (area is 10 cm 2 ); 3 cm * 5 cm (area is 15 cm 2 ). The sum of the area of the faces is 62 cm Sketch of box and possible net: c. All nets for this box have an area of 28 cm 2. d. There are two faces with an area of 8 cm 2, two with an area 2 cm 2, and two with an area of 4 cm 2 for a total of 28 cm 2. This is the same as the area of the net. 7. Figures 1, 3, 4 and 5 will not fold into a box, 2 and 6 will. Figures 2 and 6 fold to form boxes because they have edges that will match up fully and evenly when folded. The other figures will not fold to form boxes because edges that are supposed to line up with one another have different lengths, so there will either be overlaps or spaces. 8. a, b, e, f There are two of each of these faces: 2 cm * 1 cm (area is 2 cm 2 ) 2 cm * cm (area is 5 cm2 ) 1 cm * cm (area is 21 2 cm2 ). The sum of the areas of the faces is 19 cm This net will fold into an open cubic box. The two triangles will meet to become one end of the box. Covering and Surrounding 1

2 Answers 12. Sketch of box and possible net: There are six faces. Each is 5 cm * 5 cm (area is 25 cm 2 ). The sum of the areas of the faces is 150 cm a. Possible nets: There are six faces. Each is in. * 32 3 in. (each face has area in.2 ). The sum of the areas of the faces is in Sketch of box and possible net: b. Possible nets: c. Answers will depend on answers to parts (a) and (b). For the examples given above, the areas (in order) are: (a) 45 square units, 20 square units, 5 square units, and (b) 8 square units, 17 square units, and 36 square units. 15. / = 1 2 in., w = 2 in., h = 1 2 in. volume = 12 cubic 1 2 -inches volume = cubic inches surface area = in.2 Covering and Surrounding 2

3 Answers 16. / = in., w = 21 2 in., h = 11 2 in. volume = 45 cubic 1 2 -inches volume = cubic inches. surface area = in / = in., w = 21 2 in., h = 31 2 in. volume = 175 cubic 1 2 -inches volume = cubic inches surface area = in.2 Note: There are eight cubes measuring 1 2 inch on each side (8 cubic 1 2 -inches) in 1in. 3. Therefore, if you multiply the volume in cubic inches by 8, you will get the volume in cubic 1 2 -inches. 18. No, Keira does not have enough paper. The surface area of the package is 792 in. 2, which is greater than the amount of wrapping paper she has. 19. a. square b. No, you do not have enough information to find the surface area of the pyramid, because the height of the triangular faces is not known. 20. S.A. = 288 m 2 ; V = 256 m S.A. = 864 cm 2 ; V = 1,728 cm S.A. = 301 in. 2 ; V = 294 in S.A. = ft2 ; V = ft3 24. Brenda is incorrect, because the units are different for height. She could find the volume in cubic inches by multiplying 36 * 48 * 3 = 5,184 in. 3. Alternatively, she could find the volume in cubic feet by multiplying 3 * 4 * 1 4 = 3 ft a. 144 in. 3 b. 12 board-feet = 1 ft 3. Stacking 12 board-feet together would result in a stack of wood that is 1 ft by 1 ft by 12 in. c. 6 in. * 6 in. * 48 in. = 1,728 in. 3, which is greater than a board-foot (which has a volume of 144 in. 3 ). The difference between the two volumes is 1, = 1,584 in. 3. Note: If students multiply 6 * 6 * 4, they will get 144, but the unit will not be cubic inches, since they have not converted the measurement of 4 ft to 48 in. 26. a. 576 in. 3 b. Some possible dimensions include: 1 in. * 1 in. * 576 in.; 2 in. * 3 in. * 96 in.; 24 in. * 3 in. * 8 in.; 18 in. * 2 in. * 16 in.; any factor triple of 576 works. c. No, the surface areas are different. In general, objects with different dimensions typically have different surface areas. 27. a. Andrea can use 1 2 -inch blocks or 1 4 -inch blocks. b. For 1 2 -inch blocks, Andrea would need 5 blocks in a row, 7 blocks in a column, and 8 in a stack. In total, she would need inch blocks. For 1 4 -inch blocks, Andrea would need 10 blocks in a row, 14 blocks in a column, and 16 in a stack. In total, she would need 2,240 quarter-inch blocks. c. Yes, it is possible to describe the volume as 35 in A (volume of A = 35 in. 3 ; volume of B = 27 in. 3 ; volume of C = 32 in. 3 ; volume of D = 27 in. 3 ) 29. One method is to calculate the volumes of all the prisms. A second method is to notice whether or not there is one common dimension in all prisms. That way, you can find the greatest product of the dissimilar dimensions. Covering and Surrounding 3

4 Answers 30. a. 8 blocks b in. long; 3 4 in. tall; 2 in. wide c in. * 3 4 in. * 2 in. = 33 8 in.3 d. Possible solutions: 36 blocks by 1 block; 18 blocks by 2 blocks; 12 blocks by 3 blocks; 9 blocks by 4 blocks; 6 blocks by 6 blocks 31. a. Net 1: S.A square units Net 2: S.A. = 48 square units Net 3: S.A. = 32 square units Net 4: S.A. = 104 square units b. Only Nets 3 and 4 fold up to be a right rectangular prism. 32. A and E; B and F; C and D rectangles rectangles, 2 triangles square, 4 triangles in m cm ft ft 45. a. Two faces are 4 ft * 6 ft, two faces are 4 ft * 12 ft, and two faces are 6 ft * 12 ft. b. 24 ft 2 ; 48 ft 2 ; 72 ft a. Z c. 288 ft 2 b. T c. M Connections 47. A, B, C, and E all have a perimeter of 14 units; D a has perimeter of 12 units. 48. B and E 49. Any of hexominos B, C, D, or E can have one square removed to form a net for an open cubic box. Examples: 50. Hexominos B, C, D, and E can all have one square added without changing the perimeter. The perimeter does not change if you add the new square to a corner the square covers two units of perimeter while adding two new units. In the examples below, the shaded square has been added. Covering and Surrounding 4

5 Answers 51. Hexominos B, D, and E can have two squares added without changing the perimeter. For Hexominos B and D, a rectangle with a perimeter of 6 units has been added to a corner. Three units are added to the perimeter at the same time that three are being covered. Hexomino E has two units added to different corners, each unit covering two units of perimeter while adding two new units. In the examples below, the shaded squares have been added. 52. a. base = 646 feet b. 180, ft 3 c. 646 ft * 3 = 1,938 ft 9 height = feet 53. a. Both students are correct because in a right rectangular prism, the area of the base is equivalent to length times width. b. Yes, all are correct. Method 1 is the definition of surface area. Methods 2 and 3 calculate surface area and are equivalent by the Distributive Property. c. volume = in.3 surface area = in.2 Likely formulas or methods include the ones given in part (b), and V = / * w * h. d. Answers will vary. Sample answer: To find the volume, I chose to multiply length times width times height. I chose this method because it uses the three dimensions given in the Problem and I could easily substitute in the values I was given. To find the volume, I chose Method 2, S.A. = 2(/w * /w * hw). I chose this formula because, again, I could easily substitute in the values of length, width, and height. Also, it seemed faster to multiply by 2 once at the end than to multiply by 2 three times throughout the calculation, as in Method a. Matt is correct because a cube is a special type of right rectangular prism where the length, width, and height are all equal to s. b. 1,331 cm 3 c. 726 cm a. 120 in. b. They both are correct. Each edge is added 4 times to find the total edge length of the box. The two expressions are equivalent because of the Distributive Property. Extensions 56. a. The perimeter of the thickest part is 2 * * 15 = 50 in. Taking the maximum size (108 in.) and subtracting the girth (50 in.) will give the maximum height of the box = 58 in. The box can be at most 58 in. tall. b. Yes. The length is 30 in. The girth is 78 in. because = = 108, so the total size is exactly 108 in in. wide * 3 in. tall * 5 in. deep in. wide * 3 in. tall * 5 in. deep in. wide * 4 in. tall * 5 in. deep 60. Covering and Surrounding 5

6 Answers Abigail s prism is smaller; the area of the triangular base is less than the area of the rectangular base. 64. Charlie s prism is smaller, because they have the same base, but the height of Charlie s prism is less than the height of Diane s prism. 65. The base area of Elliot s prism is 24 cm 2. The base area of Fiona s prism is 24 cm 2. Their prisms have the same height, so their prisms also have the same volume a. 6 rectangles b. 2 triangles, 3 rectangles c. 2 pentagons, 5 rectangles d. 2 hexagons, 6 rectangles e. 2 decagons, 10 rectangles f. 2 n-gons, n rectangles 71. S.A. = 1 2 * 13 * 15 * 20 = 1,950 cm2 67. Covering and Surrounding 6